Equilibrium constant for a reaction can be obtained by kinetics approach or by thermodynamics approch. While from kinetics approach at equililbrium, the rate of forward and backward will be same, from thermodynamics approach Gibbs free energy will be minimized at equilibrium. Using this information and following thermodynamics values, answer the question that follow:
`DeltaG_(f)^(@)A(g)`=-200kcal/mole
`DeltaG_(f)^(@)B(g)` =-320 kcal/mole
`DeltaG_(f)^(@)C(g)`=-300kcal/mole
`DeltaG_(f)^(@)D(l)`=-224.606 kcal/mole
`DeltaG_(f)^(@)D(g)`= -226.9.9 kcal/ mole,
All values at 500K
Calculate rate constant of the backward reaction for the following reaction at 500K:
`A(g)+B(g)iffC(g)+D(l) " if " K_(f)=10"bar"^(-1) sec^(-1)`

Options

A. `10 "bar"^(-1) sec^(-1)`
B. `0.1 "bar"^(-1) sec^(-1)`
C. `0.1 sec^(-1)`
D. `10 sec^(-1)`

Answer

c